`lim_(xtooo)[x-sqrt(x^(2)-x)]` equal to ?

Options

A. `1//2`
B. 1
C. `-1//2`
D. 0

Answer

To solve the limit \( \lim_{x \to \infty} \left( x - \sqrt{x^2 - x} \right) \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to \infty} \left( x - \sqrt{x^2 - x} \right) \]

### Step 2: Factor out \( x \) from the square root
To simplify the expression under the square root, we can factor \( x^2 \) out:
\[ \sqrt{x^2 - x} = \sqrt{x^2(1 - \frac{1}{x})} = x\sqrt{1 - \frac{1}{x}} \]
Now, substitute this back into the limit:
\[ \lim_{x \to \infty} \left( x - x\sqrt{1 - \frac{1}{x}} \right) \]

### Step 3: Factor out \( x \)
Now we can factor \( x \) out of the expression:
\[ \lim_{x \to \infty} x \left( 1 - \sqrt{1 - \frac{1}{x}} \right) \]

### Step 4: Simplify the expression inside the limit
Next, we need to analyze the term \( 1 - \sqrt{1 - \frac{1}{x}} \). As \( x \to \infty \), \( \frac{1}{x} \to 0 \):
\[ 1 - \sqrt{1 - \frac{1}{x}} \approx 1 - \left( 1 - \frac{1}{2x} \right) = \frac{1}{2x} \]
Thus, we can rewrite our limit as:
\[ \lim_{x \to \infty} x \cdot \frac{1}{2x} = \lim_{x \to \infty} \frac{1}{2} = \frac{1}{2} \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to \infty} \left( x - \sqrt{x^2 - x} \right) = \frac{1}{2} \]

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